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silviumik
Pe: 2013-06-14T07:41:26+03:00 In: In:

Ajutoot trigonometrie si o ecuatie :'( !

1)Daca a,b reale astfel incat a(sin^6 x + cos^6 x) – b(sin^4 x + cos^4 x) = -1, oricare x real aratati ca a=2 si b=-3

2) Sa se rezolve ecuatia (x+a)(x+2a)(x+3a)(x+4a)=b^4 , a,b numere reale

3) Fie x E (0;pi/2) astfle incat sinx*cosx=2/4
Calculati : sin^2n x + cos^2n x, n natural

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2 raspunsuri

  1. gunty
    maestru (V)

    1.

        \[ \begin{array}{l} x = 0 \Rightarrow a\left( {0 + 1} \right) - b\left( {0 + 1} \right) = - 1 \Leftrightarrow a - b = - 1 \Leftrightarrow a = b - 1 \\ x = \frac{\pi }{4} \Rightarrow a\left[ {\left( {\frac{{\sqrt 2 }}{2}} \right)^6 + \left( {\frac{{\sqrt 2 }}{2}} \right)^6 } \right] - b\left[ {\left( {\frac{{\sqrt 2 }}{2}} \right)^4 + \left( {\frac{{\sqrt 2 }}{2}} \right)^4 } \right] = - 1 \Leftrightarrow a \cdot 2 \cdot \frac{1}{8} - b \cdot 2 \cdot \frac{1}{4} = - 1 \Leftrightarrow \frac{a}{4} - \frac{b}{2} = - 1 \Leftrightarrow a = 2b - 4 \\ \end{array} \]

    Rezolvi sistemul

        \[ \left\{ \begin{array}{l} a = b - 1 \\ a = 2b - 4 \\ \end{array} \right. \]

    si primesti a=2 si b=3.

    3.

        \[ \left. \begin{array}{l} \sin x \cdot \cos x = \frac{2}{4} \Leftrightarrow 2\sin x \cdot \cos x = 1 \Leftrightarrow \sin \left( {2x} \right) = 1 \\ x \in \left( {0,\frac{\pi }{2}} \right) \\ \end{array} \right\} \Rightarrow x = \frac{\pi }{4} \Rightarrow \sin ^{2n} x + \cos ^{2n} x = \left( {\frac{\pi }{4}} \right)^{2n} + \left( {\frac{\pi }{4}} \right)^{2n} = 2 \cdot \left( {\frac{\pi }{4}} \right)^{2n} . \]

  2. silviumik

    Multumesc de ajutor, dar daca nu te superi am gresit eu era sinx*cosx=2/5 si am incercat sa fac dupa modelul tau si nu am reusit 🙁 …

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